Negative Feedback OpAmp Circuits

A complementary push-pull transistor amplifier built exactly as shown would perform rather poorly, exhibiting crossover distortion:

The simplest way to reduce or eliminate this distortion is by adding some bias voltage to each of the transistors’ inputs, so there will never be a period of time when the two transistors are simultaneously cutoff:

One problem with this solution is that just a little too much bias voltage will result in overheating of the transistors, as they simultaneously conduct current near the zero-crossing point of the AC signal. A more sophisticated method of mitigating crossover distortion is to use an opamp with negative feedback, like this:

Explain how the opamp is able to eliminate crossover distortion in this push-pull amplifier circuit without the need for biasing.

The parasitic capacitance naturally existing in two-wire cables can cause problems when connected to high-impedance electronic devices. Take for instance certain biomedical probes used to detect electrochemical events in living tissue. Such probes may be modeled as voltage sources in series with resistances, those resistances usually being rather large due to the probes’ very small surface (contact) areas:

When connected to a cable with parasitic capacitance, a low-pass RC filter circuit is formed:

This low-pass filter (or passive integrator, if you wish) is purely unintentional. No one asked for it to be there, but it is there anyway just due to the natural resistance of the probe and the natural capacitance of the cable. Ideally, of course, we would like to be able to send the signal voltage (V_signal) straight to the amplifier with no interference or filtering of any kind so we can see exactly what it is we’re trying to measure.

One clever way of practically eliminating the effects of cable capacitance is to encase the signal wire in its own shield, and then drive that shield with the exact same amount of voltage from a voltage follower at the other end of the cable. This is called guarding:

An equivalent schematic may make this technique more understandable:

Explain why guarding the signal wire effectively eliminates the effects of the cable’s capacitance. Certainly the capacitance is still present, so how can it not have any effect on the weak signal any more?

Ideally, the input terminals of an operational amplifier conduct zero current, allowing us to simplify the analysis of many opamp circuits. However, in actuality there is a very small amount of current going through each of the input terminals of any opamp with BJT input circuitry. This may cause unexpected voltage errors in circuits.

Consider the following voltage buffer circuit:

I_bias(−) does not cause any trouble for us, because it is completely supplied by the opamp’s output. The other bias current, though, does cause trouble, because it must go through the source’s Thévenin resistance. When it does, it drops some voltage across that intrinsic source resistance, skewing the amount of voltage actually seen at the non-inverting terminal of the opamp.

A common solution to this is to add another resistor to the circuit, like this:

Explain why the addition of a resistor fixes the problem.